Graph this system of equations and solve. $y = \dfrac{3}{2} x - 1$ $y = \dfrac{7}{2} x + 3$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
The y-intercept for the first equation is $-1$ , so the first line must pass through the point $(0, -1)$ The slope for the first equation is $\dfrac{3}{2}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move up You must also move $2$ positions to the right. $2$ positions to the right. $3$ positions up from $(0, -1)$ is $(2, 2)$ Graph the blue line so it passes through $(0, -1)$ and $(2, 2)$ The y-intercept for the second equation is $3$ , so the second line must pass through the point $(0, 3)$ The slope for the second equation is $\dfrac{7}{2}$ . Remember that the slope tells you rise over run. So in this case for every $7$ positions you move up You must also move $2$ positions to the right. $2$ positions to the right. $7$ positions up from $(0, 3)$ is $(2, 10)$ Graph the green line so it passes through $(0, 3)$ and $(2, 10)$ The solution is the point where the two lines intersect. The lines intersect at $(-2, -4)$.